College Board AP 2023 (style)-style practice: Section II (long FRQ, part). A current of 2.00 A is passed through molten CuCl_2 for 30.0 minutes. Copper is deposited by Cu^2+ + 2e^- → Cu (M(Cu) = 63.5 g/mol, F = 96500\ C mol^-1). (a) Calculate the total charge passed. (b) Calculate the moles of electrons. (c) Calculate the moles of copper deposited. (d) Calculate the mass of copper deposited.

A 4-point quantitative FRQ on Faraday's law. (a) Charge (1 point): Q = It = (2.00)(30.0 × 60) = (2.00)(1800) = 3600\ C. (b) Moles of electrons (1 point): n(e^-) = QF = 360096500 = 0.0373 mol. (c) Moles of copper (1 point): the half-reaction needs 2 electrons per Cu, so n(Cu) = 0.03732 = 0.01865 mol. (d) Mass (1 point): m = n M = (0.01865)(63.5) = 1.18 g of copper. Markers reward the charge from current and time, the moles of electrons, the moles of copper using the 2-electron ratio, and the mass.

United StatesChemistrySyllabus dot point

How do we calculate the amount of substance produced at an electrode from the current and time using Faraday's law?

Topic 9.11 Electrolysis and Faraday's Law: use the current, time and the moles of electrons to calculate the mass or amount of substance produced at an electrode during electrolysis.

A focused answer to AP Chemistry Topic 9.11, covering electrolysis, the relationship between charge, current and time, Faraday's constant, and calculating the mass or moles of substance deposited or produced at an electrode, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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Quick answer

Electrolysis uses electrical energy to drive a non-spontaneous redox reaction, depositing or producing substances at the electrodes. The amount produced is found by Faraday's law in a chain of conversions. First, the total charge passed equals the current times the time: charge equals current times time, with charge in coulombs, current in amperes and time in seconds. Second, the moles of electrons equal the charge divided by Faraday's constant (96500 coulombs per mole of electrons). Third, the moles of substance produced come from the moles of electrons divided by the number of electrons in the half-reaction (for example, two electrons per copper atom). Finally, the mass is the moles times the molar mass. The half-reaction's electron count is the crucial link between electrons and atoms, so always balance the half-reaction first.

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What this topic is asking
Electrolysis and charge
From charge to moles of electrons
From electrons to mass
Try this

What this topic is asking

The College Board (Topic 9.11) wants you to use the current, time and moles of electrons to calculate the mass or amount of substance produced at an electrode during electrolysis, applying Faraday's law. This is the quantitative climax of electrochemistry, combining electrical quantities with stoichiometry.

Electrolysis and charge

The first step in any Faraday's-law problem is to find the charge. The most common slip is leaving the time in minutes or hours: convert to seconds first, because an ampere is one coulomb per second.

From charge to moles of electrons

Faraday's constant is the bridge between the electrical world (coulombs) and the chemical world (moles of electrons). Dividing the charge by $F$ converts the measured electrical charge into the moles of electrons that flowed, which is what drives the electrode reaction.

From electrons to mass

The half-reaction tells you how many electrons are needed per atom or molecule produced. Divide the moles of electrons by that number to get the moles of substance, then multiply by the molar mass for the mass:

n(\text{substance}) = \frac{n(e^-)}{\text{electrons per formula unit}}; \qquad m = n\,M

For copper, $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ needs two electrons per atom, so the moles of copper are half the moles of electrons. For silver, $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ needs one electron, so the moles of silver equal the moles of electrons. Balancing the half-reaction first is essential to get this ratio right.

It helps to see the whole calculation as a single chain of conversions, each link a familiar relationship: current and time give charge ( $Q = It$ ); charge and Faraday's constant give moles of electrons ( $n(e^-) = Q/F$ ); the half-reaction's electron count gives moles of substance; and the molar mass gives mass. Running the chain in reverse is just as valid: if a problem tells you a certain mass of metal was deposited and asks for the time or current, convert mass to moles, moles of substance to moles of electrons (using the electron count), moles of electrons to charge (multiply by $F$ ), and charge to time or current (using $Q = It$ ). Because each step is a clean multiplication or division, the only real pitfalls are forgetting to convert time to seconds and using the wrong electron count from the half-reaction.

Mass of silver deposited

A current of $1.50$ A flows for $20.0$ minutes through a silver solution, depositing silver by $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ ( $M(\text{Ag}) = 108$ g/mol).

step 1 Find the charge

$Q = It = (1.50)(20.0 \times 60) = (1.50)(1200) = 1800\ \text{C}$ .

step 2 Find the moles of electrons

$n(e^-) = \dfrac{Q}{F} = \dfrac{1800}{96500} = 0.01865$ mol.

step 3 Find the moles of silver

Silver needs one electron per atom, so $n(\text{Ag}) = 0.01865$ mol.

step 4 Find the mass

$m = n M = (0.01865)(108) = 2.01$ g of silver.

Try this

Q1. Calculate the charge passed by a $3.0$ A current in $10.$ minutes. [2 points]

Cue. $Q = It = (3.0)(10. \times 60) = (3.0)(600) = 1800\ \text{C}$ .

Q2. How many moles of aluminum are produced by $6.0$ mol of electrons, given $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ ? [2 points]

Cue. $n(\text{Al}) = \dfrac{6.0}{3} = 2.0$ mol (three electrons per aluminum atom).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A current of

2.00

A is passed through molten

\text{CuCl}_2

for

30.0

minutes. Copper is deposited by

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

(

M(\text{Cu}) = 63.5

g/mol,

F = 96500\ \text{C mol}^{-1}

). (a) Calculate the total charge passed. (b) Calculate the moles of electrons. (c) Calculate the moles of copper deposited. (d) Calculate the mass of copper deposited.

Show worked answer →

A 4-point quantitative FRQ on Faraday's law.

(a) Charge (1 point): $Q = It = (2.00)(30.0 \times 60) = (2.00)(1800) = 3600\ \text{C}$ .
(b) Moles of electrons (1 point): $n(e^-) = \dfrac{Q}{F} = \dfrac{3600}{96500} = 0.0373$ mol.
(c) Moles of copper (1 point): the half-reaction needs 2 electrons per Cu, so $n(\text{Cu}) = \dfrac{0.0373}{2} = 0.01865$ mol.
(d) Mass (1 point): $m = n M = (0.01865)(63.5) = 1.18$ g of copper.

Markers reward the charge from current and time, the moles of electrons, the moles of copper using the 2-electron ratio, and the mass.

AP 2021 (style)1 marksSection I (multiple choice). The total electric charge passed in an electrolysis is given by (A) charge equals current times time (B) charge equals current divided by time (C) charge equals current times Faraday's constant (D) charge equals time divided by current. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (A).

Charge (in coulombs) equals current (in amperes) times time (in seconds): $Q = It$ . This is the first step in any Faraday's-law calculation. The trap is (C): Faraday's constant converts charge to moles of electrons, but it is not part of the charge formula itself.

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)